The wizard has a mass of 80 kg and sits 3 m from the center of a rotating platform. Due to the rotation his speed is increased from rest by ˙ v = 0.4 m s 2 v˙=0.4ms2. If the coefficient of static friction between his clothes and the platform is μ s = 0.3 μs=0.3, determine the time requiredto cause him to slip.
Why do we equate the of both centroidal force and force in the radial direction to the friction force, but not
usR = mv^2/r?
You did not say what your acceleration was. I assume that you have both radial acceleration a= v^2/R and tangential acceleration (change in v/second).
They are perpendicular so needed friction force = m sqrt [ (v^2/R)^2 + (delta v/dt)^2 ]
Acceleration given as .4 m/s^2
then put in a = .4 m/s^2 starting at v (or more likely omega)0
I think you have typos, please check. Is it linear acceleration or angular? If angular then it is NOT m/s^2 but radians/s^2. What you gave me is gibberish.
To determine the time required for the wizard to slip, we need to consider the forces acting on him and the conditions for slipping.
The key idea is that slipping occurs when the frictional force between the wizard's clothes and the rotating platform becomes insufficient to keep him from sliding. This happens when the static friction force reaches its maximum value, given by the equation:
f_max = μ_s * N,
where μ_s is the coefficient of static friction and N is the normal force.
Now, let's analyze the forces acting on the wizard. The only force acting in the radial direction is the friction force, which points towards the center of the platform. The centripetal force, on the other hand, is provided by the net force acting on the wizard in the radial direction, given by:
F_radial = m * a_radial,
where m is the mass of the wizard and a_radial is his radial acceleration.
The centripetal force is responsible for keeping the wizard in circular motion, and it is given by:
F_centripetal = m * v^2 / r,
where v is the velocity and r is the distance of the wizard from the center of the platform.
To determine whether slipping occurs, we compare the force of static friction f_max with the radial component of the centripetal force F_radial. If f_max is greater than or equal to F_radial, slipping does not occur. However, if f_max is less than F_radial, slipping occurs.
Now, let's address your question. The equation μ_s * R = m * v^2 / r is not used to determine slipping because it relates the static friction force to the centripetal force. This equation assumes that the static friction is the only force counteracting the centripetal force, which is not the case here. There is another external force acting on the wizard - the force that increases his speed, given by ˙ v = 0.4 m s^(-2). This force produces an additional acceleration and must also be considered when determining the slipping condition.
Thus, to determine whether slipping occurs, we need to compare the static friction force f_max with the radial component of the net force F_net in the radial direction. The net force is the vector sum of the centripetal force F_centripetal and the force that increases the wizard's speed. Only by comparing f_max with F_net can we determine whether slipping occurs.
I hope this explanation clarifies why we do not use the equation μ_s * R = m * v^2 / r to determine slipping in this particular scenario.